Detailed Solution

1. Key Concept: Binomial Theorem

The core concept for solving this problem is the Binomial Theorem, which provides a formula for expanding expressions of the form $(a+b)^n$. The general term, also known as the $(r+1)^{th}$ term, in the expansion of $(a+b)^n$ is given by: $T_{r+1} = {}^nC_r \cdot a^{n-r} \cdot b^r$ where:

• $n$ is the power to which the binomial is raised.
• $r$ is the index of the term, starting from 0 for the first term.
• ${}^nC_r$ is the binomial coefficient, calculated as ${n! \over r!(n-r)!}$.
• $a$ is the first term of the binomial.
• $b$ is the second term of the binomial.

2. Identifying Terms and Applying the Formula

We are given the binomial expansion of ${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$. By comparing this with $(a+b)^n$, we can identify the components:

• $a = 1$
• $b = {x^{{{\log }_2}x}}$
• $n = 5$

We need to find the third term, which means $T_3$. For the formula $T_{r+1}$, if $r+1 = 3$, then $r = 2$.

Now, substitute these values into the general term formula: $T_3 = T_{2+1} = {}^5C_2 \cdot (1)^{5-2} \cdot \left( {x^{{{\log }_2}x}} \right)^2$

3. Calculating the Third Term

Let's simplify the expression for $T_3$ step-by-step:

• Calculate the binomial coefficient ${}^5C_2$: ${}^5C_2 = {5! \over 2!(5-2)!} = {5! \over 2!3!} = {{5 \times 4 \times 3 \times 2 \times 1} \over {(2 \times 1) \times (3 \times 2 \times 1)}} = {{5 \times 4} \over {2 \times 1}} = 10$ Explanation: This calculates the number of ways to choose 2 items from a set of 5, which is the coefficient for the third term.

• Simplify the term with base 1: $(1)^{5-2} = (1)^3 = 1$ Explanation: Any power of 1 is always 1.

• Simplify the exponential term: ${\left( {{x^{{{\log }_2}x}}} \right)^2}$ Using the exponent rule $(A^B)^C = A^{B \cdot C}$: ${\left( {{x^{{{\log }_2}x}}} \right)^2} = {x^{2 \cdot {{\log }_2}x}}$ Explanation: When raising a power to another power, we multiply the exponents.

Combining these, the third term $T_3$ is: $T_3 = 10 \cdot 1 \cdot {x^{2{{\log }_2}x}}$ $T_3 = 10 \cdot {x^{2{{\log }_2}x}}$

4. Forming the Equation

The problem states that the third term equals 2560. So, we set up the equation: $10 \cdot {x^{2{{\log }_2}x}} = 2560$ To simplify, divide both sides by 10: ${x^{2{{\log }_2}x}} = 256$ Explanation: This isolates the exponential term involving $x$, preparing it for further logarithmic manipulation.

5. Solving for x using Logarithms

When the variable appears in both the base and the exponent, taking a logarithm of both sides is a common strategy to bring the exponent down. Since the exponent already involves $\log_2 x$, taking $\log_2$ on both sides will simplify the equation.

• Take $\log_2$ on both sides: $\log_2 \left( {x^{2{{\log }_2}x}} \right) = \log_2 (256)$ Explanation: Applying the logarithm function to both sides maintains the equality and allows us to use logarithm properties to simplify the left side. We choose base 2 because it matches the base of the logarithm already present in the exponent, which will lead to significant simplification.

• Apply the logarithm property $\log_b (A^C) = C \cdot \log_b A$: The left side becomes: $(2{{\log }_2}x) \cdot ({{\log }_2}x)$ The right side: We need to express 256 as a power of 2. $256 = 2^8$ So, $\log_2 (256) = \log_2 (2^8) = 8$ Explanation: This step brings the complex exponent down to the base line, transforming the equation into a more manageable form. Simplifying $\log_2 256$ makes the right side a simple constant.

• Substitute these back into the equation: $2 \cdot ({{\log }_2}x) \cdot ({{\log }_2}x) = 8$ $2 ({{\log }_2}x)^2 = 8$

• Simplify and solve for ${{\log }_2}x$: Divide both sides by 2: $({{\log }_2}x)^2 = 4$ Take the square root of both sides. Remember that taking the square root can yield both a positive and a negative result: ${{\log }_2}x = \pm \sqrt{4}$ ${{\log }_2}x = \pm 2$ Explanation: We isolate the term ${{\log }_2}x$ by performing algebraic operations. It's crucial to remember both the positive and negative roots when solving an equation of the form $y^2 = k$.

6. Finding Possible Values of x

Now we have two possible cases for ${{\log }_2}x$:

• Case 1: ${{\log }_2}x = 2$ Using the definition of logarithm ($\log_b A = C \iff A = b^C$): $x = 2^2$ $x = 4$

• Case 2: ${{\log }_2}x = -2$ Using the definition of logarithm: $x = 2^{-2}$ $x = {1 \over {2^2}}$ $x = {1 \over 4}$ Explanation: We convert the logarithmic equations back into exponential form to find the values of $x$. Both values are positive, satisfying the domain requirement for $\log_2 x$ ($x>0$).

Thus, the possible values of $x$ are 4 and $1 \over 4$. From the given options, $1 \over 4$ matches option (D).

7. Common Mistakes & Tips

• Forgetting the $\pm$ sign: A common error is only considering the positive square root when solving $y^2 = k$. Always include both positive and negative roots.
• Incorrect Logarithm Properties: Be careful with logarithm rules. For example, $({\log_b x})^2$ is not the same as $\log_b (x^2)$.
• Domain of Logarithms: Always ensure that the arguments of logarithms are positive. In this case, for $\log_2 x$ to be defined, $x > 0$. Both our solutions (4 and $1/4$) satisfy this condition.
• Base Consistency: When taking logarithms on both sides, choose a base that simplifies the equation, typically matching a base already present in the expression.

8. Summary/Key Takeaway

This problem effectively tests the understanding of the Binomial Theorem's general term formula and the fundamental properties of logarithms. The key steps involved calculating the binomial coefficient, simplifying exponential terms, applying logarithms to solve for an unknown in an exponent, and correctly handling both positive and negative solutions from a quadratic-like equation. The possible values of $x$ that satisfy the given condition are 4 and $1 \over 4$.